We saw in Sec. 9.3 that TEM modes are described by Eqs. (9.3.3) and (9.3.4), the latter being equivalent to a two-dimensional electrostatic problem:
HT= 1 ηˆz×ET
∇∇∇T×ET=0
∇∇∇TãET=0
(TEM modes) (10.1.1)
The second of (10.1.1) implies thatETcan be expressed as the (two-dimensional) gradient of a scalar electrostatic potential. Then, the third equation becomes Laplace’s equation for the potential. Thus, the electric field can be obtained from:
∇2Tϕ=0 ET= −∇∇∇Tϕ
(equivalent electrostatic problem) (10.1.2)
Because in electrostatic problems the electric field lines must start at positively charged conductors and end at negatively charged ones, a TEM mode can be supported only in multi-conductor guides, such as the coaxial cable or the two-wire line. Hollow conducting waveguides cannot support TEM modes.
Fig. 10.1.1 depicts the transverse cross-sectional area of a two-conductor transmis- sion line. The cross-section shapes are arbitrary.
The conductors areequipotentials of the electrostatic solution. Letϕa, ϕbbe the constant potentials on the two conductors. The voltage difference between the conduc- tors will beV=ϕa−ϕb. The electric field lines start perpendicularly on conductor (a) and end perpendicularly on conductor (b).
The magnetic field lines, being perpendicular to the electric lines according to Eq. (10.1.1), are recognized to be theequipotential lines. As such, they close upon themselves sur- rounding the two conductors.
398 10. Transmission Lines
Fig. 10.1.1 Two-conductor transmission line.
In particular, on the conductor surfaces the magnetic field is tangential. According to Amp`ere’s law, the line integrals of the magnetic field around each conductor will result into total currentsIand−Iflowing on the conductors in thez-direction. These currents are equal and opposite.
Impedance, Inductance, and Capacitance
Because the fields are propagating along thez-direction with frequencyωand wavenum- berβ=ω/c, thez, tdependence of the voltageVand currentIwill be:
V(z, t)=Vejωt−jβz
I(z, t)=Iejωt−jβz (10.1.3) For backward-moving voltage and current waves, we must replaceβby−β. The ratio V(z, t)/I(z, t)=V/Iremains constant and independent ofz. It is called thecharacter- istic impedanceof the line:
Z=V
I (line impedance) (10.1.4)
In addition to the impedanceZ, a TEM line is characterized by its inductance per unit lengthLand its capacitance per unit lengthC. For lossless lines, the three quantities Z, L, Care related as follows:
L=μZ
η, C=η
Z (inductance and capacitance per unit length) (10.1.5) whereη=
μ/is the characteristic impedance of the dielectric medium between the conductors.†By multiplying and dividingLandC, we also obtain:
Z=
L
C, c=√1μ=√1
LC (10.1.6)
†These expressions explain whyμandare sometimes given in units of henry/m and farad/m.
10.1. General Properties of TEM Transmission Lines 399 Thevelocity factorof the line is the ratioc/c0=1/n, wheren=
/0=√ ris the refractive index of the dielectric, which is assumed to be non-magnetic.
Becauseω=βc, the guide wavelength will beλ=2π/β=c/f =c0/f n=λ0/n, whereλ0is the free-space wavelength. For a finite lengthlof the transmission line, the quantityl/λ=nl/λ0is referred to as theelectrical lengthof the line and plays the same role as the optical length in thin-film layers.
Eqs. (10.1.5) and (10.1.6) are general results that are valid for any TEM line. They can be derived with the help of Fig. 10.1.2.
Fig. 10.1.2 Surface charge and magnetic flux linkage.
The voltageVis obtained by integratingETãdlalong any path from (a) to (b). How- ever, if that path is chosen to be anE-field line, thenETãdl= |ET|dl, giving:
V= b
a|ET|dl (10.1.7)
Similarly, the currentIcan be obtained by the integral ofHTãdlalong any closed path around conductor (a). If that path is chosen to be anH-field line, such as the peripheryCaof the conductor, we will obtain:
I=
Ca
|HT|dl (10.1.8)
The surface charge accumulated on an infinitesimal areadl dzof conductor (a) is dQ = ρsdl dz, whereρs is the surface charge density. Because the conductors are assumed to be perfect, the boundary conditions require thatρsbe equal to the normal component of theD-field, that is,ρs=|ET|. Thus,dQ=|ET|dl dz.
If we integrate over the peripheryCaof conductor (a), we will obtain the total surface charge per unitz-length:
Q=dQ dz =
Ca
|ET|dl
But because of the relationship|ET| =η|HT|, which follows from the first of Eqs. (10.1.1), we have:
Q=
Ca
|ET|dl=η
Ca
|HT|dl=ηI (10.1.9) where we used Eq. (10.1.8). BecauseQis related to the capacitance per unit length and the voltage byQ=CV, we obtain
400 10. Transmission Lines
Q=CV=ηI ⇒ C=η I V =η
Z
Next, we consider anE-field line between pointsAandBon the two conductors. The magnetic flux through the infinitesimal areadl dzwill bedΦ= |BT|dl dz=μ|HT|dl dz because the vectorHTis perpendicular to the area.
If we integrate from (a) to (b), we will obtain the total magnetic flux linking the two conductors per unitz-length:
Φ=dΦ dz =
b a
μ|HT|dl replacing|HT| = |ET|/ηand using Eq. (10.1.7), we find:
Φ= b
aμ|HT|dl=μ η
b
a|ET|dl=μ ηV
The magnetic flux is related to the inductance via Φ=LI. Therefore, we get:
Φ=LI=μ
ηV ⇒ L=μ η
V I =μZ
η Transmitted Power
The relationships amongZ, L, Ccan also be derived using energy considerations. The power transmitted along the line is obtained by integrating the z-component of the Poynting vector over the cross-sectionS of the line. For TEM modes we havePz =
|ET|2/2η, therefore, PT=21
η
S|ET|2dx dy=21 η
S|∇∇∇Tϕ|2dx dy (10.1.10) It can be shown in general that Eq. (10.1.10) can be rewritten as:
PT=1
2Re(V∗I)=1
2Z|I|2= 1
2Z|V|2 (10.1.11)
We will verify this in the various examples below. It can be proved using the following Green’s identity:
|∇∇∇Tϕ|2+ϕ∗∇2Tϕ= ∇∇∇Tã(ϕ∗∇∇∇Tϕ) WritingET= −∇∇∇Tϕand noting that∇2Tϕ=0, we obtain:
|ET|2= −∇∇∇Tã(ϕ∗ET) Then, the two-dimensional Gauss’ theorem implies:
10.1. General Properties of TEM Transmission Lines 401
PT= 1 2η
S|ET|2dx dy= −1 2η
S∇∇∇Tã(ϕ∗ET)dx dy
= − 1 2η
Ca
ϕ∗ETã(−ˆn)dl− 1 2η
Cb
ϕ∗ETã(−ˆn)dl
= 1 2η
Ca
ϕ∗(ETãnˆ) dl+ 1 2η
Cb
ϕ∗(ETãnˆ) dl
where ˆnare the outward normals to the conductors (the quantity−ˆnis the normal outward from the regionS.) Because the conductors are equipotential surfaces, we have ϕ∗=ϕ∗aon conductor (a) andϕ∗=ϕ∗bon conductor (b). Using Eq. (10.1.9) and noting thatETãˆn= ±|ET|on conductors (a) and (b), we obtain:
PT= 1 2ηϕ∗a
Ca
|ET|dl− 1 2ηϕ∗b
Cb
|ET|dl= 1 2ηϕ∗aQ
− 1 2ηϕ∗bQ
=1
2(ϕ∗a−ϕ∗b)Q η=1
2V∗ηI η =1
2V∗I=1 2Z|I|2
The distribution of electromagnetic energy along the line is described by the time- averaged electric and magnetic energy densities per unit length, which are given by:
We=1 4
S|ET|2dx dy , Wm=1 4μ
S|HT|2dx dy Using Eq. (10.1.10), we may rewrite:
We=12ηPT=21
cPT, Wm=12μ ηPT=21
cPT
Thus,We =Wmand the total energy density isW = We+Wm =PT/c, which implies that the energy velocity will beven =PT/W = c. We may also express the energy densities in terms of the capacitance and inductance of the line:
We=1
4C|V|2, Wm=1
4L|I|2 (10.1.12)
Power Losses, Resistance, and Conductance
Transmission line losses can be handled in the manner discussed in Sec. 9.2. The field patterns and characteristic impedance are determined assuming the conductors are per- fectly conducting. Then, the losses due to the ohmic heating of the dielectric and the conductors can be calculated by Eqs. (9.2.5) and (9.2.9).
These losses can be quantified by two more characteristic parameters of the line, the resistance and conductance per unit length,RandG. The attenuation coefficients due to conductor and dielectric losses are then expressible in termsR,GandZby:
αc= R
2Z, αd=1
2GZ (10.1.13)
402 10. Transmission Lines
They can be derived in general terms as follows. The induced surface currents on the conductor walls areJs=ˆn×HT=ˆn×(ˆz×ET)/η, where ˆnis the outward normal to the wall.
Using the BAC-CAB rule, we findJs =ˆz(nˆãET)/η. But, ˆnis parallel toETon the surface of conductor (a), and anti parallel on (b). Therefore, ˆnãET= ±|ET|. It follows thatJs= ±ˆz|ET|/η= ±ˆz|HT|, pointing in the+zdirection on (a) and−zdirection on (b). Inserting these expressions into Eq. (9.2.8), we find for the conductor power loss per unitz-length:
Ploss=dPloss
dz =1 2Rs
Ca
|HT|2dl+1 2Rs
Cb
|HT|2dl (10.1.14) BecauseHTis related to the total currentIvia Eq. (10.1.8), we may define the resis- tance per unit lengthRthrough the relationship:
Ploss=12R|I|2 (conductor ohmic losses) (10.1.15) Using Eq. (10.1.11), we find for the attenuation coefficient:
αc=Ploss 2PT =
1 2R|I|2 21
2Z|I|2 = R
2Z (10.1.16)
If the dielectric between the conductors is slightly conducting with conductivityσd
or loss tangent tanδ=σd/ω, then there will be some current flow between the two conductors.
The induced shunt current per unitz-length is related to the conductance byId= GV. The shunt current density within the dielectric isJd = σdET. The total shunt current flowing out of conductor (a) towards conductor (b) is obtained by integratingJd around the periphery of conductor (a):
Id=
Ca
Jdãˆndl=σd
Ca
|ET|dl Using Eq. (10.1.9), we find:
Id=σdQ
=GV ⇒ G=σd
C=σdη Z It follows that the dielectric loss constant (9.2.5) will be:
αd=12σdη=12GZ
Alternatively, the power loss per unit length due to the shunt current will bePd= Re(IdV∗)/2=G|V|2/2, and therefore,αdcan be computed from:
αd= Pd 2PT=
1 2G|V|2 2 1
2Z|V|2 =12GZ